- #1

#### PaultheRiemann

A \sqrt{x^2 + y^2}

The double integral

\iint A \sqrt{x^2 + y^2} \space dy \space dx \space \space \space

\text {From x= -1 to 1 and y=} -\sqrt{1-x^2} \space to \space \sqrt{1-x^2}

\text{ is very difficult to evaluate. I've tried polar coordinate substitution. However, I can't seem to get} /space 2/3 \pi z \space \text{where z is the height of cone and 1 is the radius of the cone}